PYQs / GATE CE / 2020 / Set 2 / Q49 GATE CE 2020 Set 2 — Question 49 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. NAT +2 / -0 Medium Fourier Series Partial Differential Equations Engineering Mathematics
Engineering Mathematics → Partial Differential Equations → Fourier Series
Last updated 5 September 2026
Question The Fourier series to represent
x − x 2 x - x^2 x − x 2 for
− π ≤ x ≤ π -\pi \leq x \leq \pi − π ≤ x ≤ π is given by
x − x 2 = a 0 2 + ∑ n = 1 ∞ a n cos n x + ∑ n = 1 ∞ b n sin n x x - x^2 = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos nx + \sum_{n=1}^{\infty} b_n \sin nx x − x 2 = 2 a 0 + n = 1 ∑ ∞ a n cos n x + n = 1 ∑ ∞ b n sin n x The value of
a 0 a_0 a 0 (round off to two decimal places), is
________ .
Solution The coefficient
a 0 a_0 a 0 in a Fourier series for a function
f ( x ) f(x) f ( x ) on
[ − π , π ] [-\pi, \pi] [ − π , π ] is defined as:
a 0 = 1 π ∫ − π π f ( x ) d x a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx a 0 = π 1 ∫ − π π f ( x ) d x Here,
f ( x ) = x − x 2 f(x) = x - x^2 f ( x ) = x − x 2 .
a 0 = 1 π ∫ − π π ( x − x 2 ) d x a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} (x - x^2) dx a 0 = π 1 ∫ − π π ( x − x 2 ) d x Evaluating the integral:
∫ − π π x d x = 0 (since x is an odd function) \int_{-\pi}^{\pi} x dx = 0 \text{ (since } x \text{ is an odd function)} ∫ − π π x d x = 0 (since x is an odd function) ∫ − π π − x 2 d x = − 2 ∫ 0 π x 2 d x = − 2 [ x 3 3 ] 0 π = − 2 π 3 3 \int_{-\pi}^{\pi} -x^2 dx = -2 \int_{0}^{\pi} x^2 dx = -2 \left[ \frac{x^3}{3} \right]_0^\pi = -\frac{2\pi^3}{3} ∫ − π π − x 2 d x = − 2 ∫ 0 π x 2 d x = − 2 [ 3 x 3 ] 0 π = − 3 2 π 3 So,
a 0 = 1 π ( − 2 π 3 3 ) = − 2 π 2 3 a_0 = \frac{1}{\pi} \left( -\frac{2\pi^3}{3} \right) = -\frac{2\pi^2}{3} a 0 = π 1 ( − 3 2 π 3 ) = − 3 2 π 2 Using
π ≈ 3.14159 \pi \approx 3.14159 π ≈ 3.14159 :
a 0 = − 2 × ( 3.14159 ) 2 3 ≈ − 2 × 9.8696 3 ≈ − 6.5797 a_0 = -\frac{2 \times (3.14159)^2}{3} \approx -\frac{2 \times 9.8696}{3} \approx -6.5797 a 0 = − 3 2 × ( 3.14159 ) 2 ≈ − 3 2 × 9.8696 ≈ − 6.5797 Rounding to two decimal places,
a 0 = − 6.58 a_0 = -6.58 a 0 = − 6.58 .
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Solution The coefficient
a 0 a_0 a 0 in a Fourier series for a function
f ( x ) f(x) f ( x ) on
[ − π , π ] [-\pi, \pi] [ − π , π ] is defined as:
a 0 = 1 π ∫ − π π f ( x ) d x a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx a 0 = π 1 ∫ − π π f ( x ) d x Here,
f ( x ) = x − x 2 f(x) = x - x^2 f ( x ) = x − x 2 .
a 0 = 1 π ∫ − π π ( x − x 2 ) d x a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} (x - x^2) dx a 0 = π 1 ∫ − π π ( x − x 2 ) d x Evaluating the integral:
∫ − π π x d x = 0 (since x is an odd function) \int_{-\pi}^{\pi} x dx = 0 \text{ (since } x \text{ is an odd function)} ∫ − π π x d x = 0 (since x is an odd function) ∫ − π π − x 2 d x = − 2 ∫ 0 π x 2 d x = − 2 [ x 3 3 ] 0 π = − 2 π 3 3 \int_{-\pi}^{\pi} -x^2 dx = -2 \int_{0}^{\pi} x^2 dx = -2 \left[ \frac{x^3}{3} \right]_0^\pi = -\frac{2\pi^3}{3} ∫ − π π − x 2 d x = − 2 ∫ 0 π x 2 d x = − 2 [ 3 x 3 ] 0 π = − 3 2 π 3 So,
a 0 = 1 π ( − 2 π 3 3 ) = − 2 π 2 3 a_0 = \frac{1}{\pi} \left( -\frac{2\pi^3}{3} \right) = -\frac{2\pi^2}{3} a 0 = π 1 ( − 3 2 π 3 ) = − 3 2 π 2 Using
π ≈ 3.14159 \pi \approx 3.14159 π ≈ 3.14159 :
a 0 = − 2 × ( 3.14159 ) 2 3 ≈ − 2 × 9.8696 3 ≈ − 6.5797 a_0 = -\frac{2 \times (3.14159)^2}{3} \approx -\frac{2 \times 9.8696}{3} \approx -6.5797 a 0 = − 3 2 × ( 3.14159 ) 2 ≈ − 3 2 × 9.8696 ≈ − 6.5797 Rounding to two decimal places,
a 0 = − 6.58 a_0 = -6.58 a 0 = − 6.58 .
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More questions on Partial Differential Equations ← Q48 Full paper Q50 → 